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Math Help - Group of order pq must have subgroups of order p and q.

  1. #16
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    Quote Originally Posted by Swlabr View Post
    Chauchy's Theorem is used to prove that finite groups with elements of order a power of p are of order p^n. However, in this case the elements are of order p. This is a simpler case, and we can use the formula,

    |HK|=|H|.|K|/|H \cap K|

    That is to say, if we look at the problem in my proof each element has order p. As each element has order p it will generate p-2 other elements of order p and so if we take a minimum generating set, P_1, P_2, \ldots, P_n, the intersection of each pair is trivial, G=P_1P_2 \ldots P_n and the formula can be applied. Thus,

    |G|=|P_1|.|P_2| \ldots |P_n| = p^n \neq pq

    and we have the contradiction needed to salvage my earlier proof.

    I hope that makes sense...

    It should be noted that that formula has an elementary proof, and so I believe that my proof will count as "elementary". I hope.
    Are you telling me that whenever I have subgroups H_1,H_2, \ldots,H_m of G such that H_i \cap H_j = (e) for i \neq j, |H_1H_2 \ldots H_m| = |H_1||H_2| \ldots |H_m|? This is certainly true for m=2. Could you sugest a proof for m any positive integer?
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  2. #17
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    Are you telling me that whenever I have subgroups H_1,H_2, \ldots,H_m of G such that H_i \cap H_j = (e) for i \neq j, |H_1H_2 \ldots H_m| = |H_1||H_2| \ldots |H_m|? This is certainly true for m=2. Could you sugest a proof for m any positive integer?
    No, I am not. For instance, take the Klein 4-group, V_4=C_2 \times C_2. We have 4 copies of C_2 in this group with trivial pairwise intersections. However, their product is V_4, a group of order 4 not of order 8. We need two more things - that the subgroups are of prime order, and that they form a minimal generating set. That is to say, H_1H_2 \ldots H_m=G but H_1H_2 \ldots H_{i-1}H_{i+1} \ldots H_m \neq G.

    The flaw in the counter example is that they do not form a minimal generating set. We need prime order as we do not know that AB \cap C = <1> just because A \cap C = B \cap C = <1> even if we do have a minimum generating set.

    As to the proof - think...what would you normally do when something is true for m=2 (actually - easier than that - it's true for m=1)? You will need the formula I gave though, as well as the two conditions I've given, above.
    Last edited by Swlabr; September 15th 2009 at 06:32 AM.
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  3. #18
    MHF Contributor Swlabr's Avatar
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    I have another question regarding such groups that I started to ponder but could not really get anywhere on...are all such groups soluble? If so, then we are done as either they are abelian and so we have that they are isomorphic to C_{pq} \equiv C_p \times C_q, or else the derived subgroup is a proper normal subggroup which we can quotient out and then apply the correspondence theorem (I believe we can anyway) to get the other subgroup.

    However, I can't seem to prove that all such groups are soluble, but I can't find any counter-examples (admittedly my examples were far from many in number...emm, A_6 was all I could really think of). Clearly there are no non-abelian nilpotent examples, as their centres are trivial.
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  4. #19
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    Quote Originally Posted by Swlabr View Post
    I have another question regarding such groups that I started to ponder but could not really get anywhere on...are all such groups soluble?
    Yes, they are all solvable. You can easily see this from Burnside's theorem but this is a complete overkill.
    This special case can also be proved with the Sylow theorems.
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  5. #20
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    Quote Originally Posted by Swlabr View Post
    As to the proof - think...what would you normally do when something is true for m=2 (actually - easier than that - it's true for m=1)? You will need the formula I gave though, as well as the two conditions I've given, above.
    I would use induction, of course. I think I have a proof, and it would run as follows: suppose the proposition to be true for groups admitting a decomposition in m subgroups (hypothesis) and that G=H_1 \ldots H_{m+1} is also a decomposition satisfying the three conditions. Then G=HH_{m+1} with H=H_1 \ldots H_m. Obviously H satisfies the three conditions. Now, lets suppose for a moment that two things happen:
    (a) H is a subgroup of G.
    (b) H \cap H_{m+1} = (e).
    Then, by the hypothesis, as H is a group in its own right, |H| = |H_1| \ldots |H_m|. And
    by applying the formula, I get |G| = |H||H_{m+1}| and the proposition is proved. Unfortunately, I don't see how (a) and (b) can be proved.
    Last edited by ENRIQUESTEFANINI; September 15th 2009 at 12:10 PM. Reason: Something missing.
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